Original model

The original model describes steady, one-dimensional flow along the mean streamline of an annular channel. The governing equations are:

vmdρdm+ρdvmdm=ρvmAdAdmρvmdvmdm+dpdm=ρvθ2rsin(ϕ)2τbcos(α)ρvmdvθdm=ρvθvmrsin(ϕ)2τbsin(α)ρvmdpdmρvma2dρdm=2(τv+q˙)b(ep)ρ\begin{gather} v_m \frac{d\rho}{dm} + \rho \frac{dv_m}{dm} = -\frac{\rho v_m}{A} \frac{dA}{dm} \\[4pt] \rho v_m \frac{dv_m}{dm} + \frac{dp}{dm} = \frac{\rho v_\theta^2}{r} \sin(\phi) - \frac{2 \tau}{b} \cos(\alpha) \\[4pt] \rho v_m \frac{dv_\theta}{dm} = -\frac{\rho v_\theta v_m}{r} \sin(\phi) - \frac{2 \tau}{b} \sin(\alpha) \\[4pt] \rho v_m \frac{dp}{dm} - \rho v_m a^2 \frac{d\rho}{dm} = \frac{2(\tau v + \dot{q})}{b \left( \frac{\partial e}{\partial p} \right)_\rho} \end{gather}

The first equation expresses the mass balance in the annular passage. The second and third equations represent the momentum balance in the meridional and tangential directions, respectively. The last equation represents thermal energy balance, incorporating viscous dissipation and wall heat transfer. The viscous shear stress is modeled using a conventional friction factor approach:

τ=12cfρv2\tau = \frac{1}{2} c_f \rho v^2

where cfc_f is the skin friction coefficient, obtained from empirical correlations as a function of the Reynolds number. Thermodynamic properties are evaluated from the equation of state, using pressure–density as input pair.

"R. Agromayor, B. Müller, and L. O. Nord, “One-dimensional annular diffuser model for preliminary turbomachinery design,” International Journal of Turbomachinery, Propulsion and Power, vol. 4, no. 3, p. 31, 2019, doi: 10.3390/ijtpp4030031.

New model

An equivalent formulation of the model can be derived from the total energy equation, leading to the following system of governing equations:

ρdvmdm+vmdρdm=ρvmAdAdmρvmdvmdm+dpdm=ρvθ2rsin(ϕ)2τbcos(α)ρvmdvθdm=ρvθvmrsin(ϕ)2τbsin(α)vmdvmdm+vθdvθdm+dhdm=2q˙b\begin{gather} \rho \frac{dv_m}{dm} + v_m \frac{d \rho}{dm} = -\frac{\rho v_m}{A} \frac{dA}{dm} \\ \rho v_m \frac{dv_m}{dm} + \frac{dp}{dm} = \frac{\rho v_\theta^2}{r} \sin(\phi) - \frac{2 \tau}{b} \cos(\alpha) \\ \rho v_m \frac{dv_\theta}{dm} = -\frac{\rho v_\theta v_m}{r} \sin(\phi) - \frac{2 \tau}{b} \sin(\alpha) \\ v_m \frac{dv_m}{dm} + v_\theta \frac{dv_\theta}{dm} + \frac{d h}{dm} = \frac{2 \dot{q}}{b} \end{gather}

To close the system, the density is expressed as a function of pressure and enthalpy:

dρdm=(ρp)hdpdm+vm(ρh)pdhdm \frac{d \rho}{dm} = \left( \frac{\partial \rho}{\partial p} \right)_h \frac{dp}{dm} + v_m \left( \frac{\partial \rho}{\partial h} \right)_p \frac{d h}{dm}

where the partial derivatives of density can be expressed in terms of the speed of sound and the Gruneisen parameter:

(ρh)p=ρGc2(ρp)h=1+Gc2\begin{align} \left(\frac{\partial \rho}{\partial h}\right)_{p} = -\frac{\rho G}{c^2} \\[0.5em] \left(\frac{\partial \rho}{\partial p}\right)_{h} = \frac{1 + G}{c^2} \end{align}

Substituting the above thermodynamic relations into the continuity equation gives the following quasi-linear system of ordinary differential equations:

ρvmdvmdm+1+Gc2dpdmρGc2dhdm=ρAdAdmρvmdvmdm+dpdm=ρvθ2rsin(ϕ)2τbcos(α)ρvmdvθdm=ρvθvmrsin(ϕ)2τbsin(α)vmdvmdm+vθdvθdm+dhdm=2q˙b\begin{gather} \frac{\rho}{v_m} \frac{dv_m}{dm} + \frac{1 + G}{c^2} \frac{dp}{dm} - \frac{\rho G}{c^2} \frac{d h}{dm} = -\frac{\rho}{A} \frac{d A}{dm} \\ \rho v_m \frac{dv_m}{dm} + \frac{dp}{dm} = \frac{\rho v_\theta^2}{r} \sin(\phi) - \frac{2 \tau}{b} \cos(\alpha) \\ \rho v_m \frac{dv_\theta}{dm} = -\frac{\rho v_\theta v_m}{r} \sin(\phi) - \frac{2 \tau}{b} \sin(\alpha) \\ v_m \frac{dv_m}{dm} + v_\theta \frac{dv_\theta}{dm} + \frac{d h}{dm} = \frac{2 \dot{q}}{b} \end{gather}

Characteristic determinant and singularity condition

This system can be expressed in compact matrix form as:

AdUdm  =  SA\,\frac{dU}{dm} \;=\; S
A=[ρvm0ρGc21+Gc2ρvm0010ρvm00vmvθ10],U=[vmvθhp],S=[ρAdAdmρvθ2rsinϕ2τbcosαρvθvmrsinϕ2τbsinα2q˙wb].A = \begin{bmatrix} \tfrac{\rho}{v_m} & 0 & -\,\tfrac{\rho G}{c^{2}} & \tfrac{1+G}{c^{2}} \\ \rho v_m & 0 & 0 & 1 \\ 0 & \rho v_m & 0 & 0 \\ v_m & v_\theta & 1 & 0 \end{bmatrix}, \qquad U = \begin{bmatrix} v_m\\ v_\theta\\ h\\ p \end{bmatrix}, \qquad S = \begin{bmatrix} -\tfrac{\rho}{A}\,\tfrac{dA}{dm} \\ \tfrac{\rho v_\theta^{2}}{r}\,\sin\phi - \tfrac{2\tau}{b}\cos\alpha \\ -\tfrac{\rho v_\theta v_m}{r}\,\sin\phi - \tfrac{2\tau}{b}\sin\alpha \\ \tfrac{2\,\dot q_w}{b} \end{bmatrix}.

The determinant of the coefficient matrix AA is:

det(A)=ρ2(1Mam2),Mam=vma\det(A) = \rho^{2} \left(1 - Ma_m^{2}\right), \qquad Ma_m = \frac{v_m}{a}

This result shows that the system becomes singular when the meridional Mach number equals unity. At this condition, the flow reaches a sonic state along the mean streamline, corresponding to choking at the section of minimum area.

Total pressure and entropy generation

The mechanical energy equation with this term is given by:

ρvm(1ρdpdm+vdvdm)=vmdp0dm=2bτvIvm\rho v_m \left( \frac{1}{\rho}\frac{dp}{dm} + v \frac{dv}{dm} \right) = v_m \frac{dp_0}{dm} = -\frac{2}{b}\tau v - I v_m
dp0dm=2τbcosαI\frac{dp_0}{dm} = -\frac{2\tau }{b \cos \alpha} - I

This equation makes a lot of sense. The decrease in stagnation pressure is caused by the wall friction, plus any other sink terms that we put in the momentum equations. In this case we see clearly that I has the physical meaning of stagnation pressure drop.

Their effect on the entropy generation expression is given by

σT=2b(1TTw)qw+2bτv+Ivm\sigma T = \frac{2}{b}\left(1 - \frac{T}{T_w} \right)q_w + \frac{2}{b} \tau v + I v_m

In general, which even term appears on the right hand side of the momentum equations, but does not produce work in the energy equation is a loss mechanism that result in a loss of total pressure and generation of entropy

Diffusion and curvature losses

Viscous losses can be modeled as a contribution of 3 factors:

There 3 contributions can be lumped together into the shear stress term τ\tau, which affects both the meridional and tangential momentum components. For some reason, Aungier puts all the loss due to diffusion and curvature into the meridional component. I do not think that there is a strong physical basis for this, and I believe tat this will cause distortions in the exit flow angle.

In order to understand how to model the losses (i.e., loss of total stagnation pressure) we first need to derive the equation for stagnation pressure. Multiplying the momentum equations by their respective velocity components

ρvmddm(vm22)+vmdpdm=ρvθ2vmrsin(ϕ)2τbvmcos(α)ρvmddm(vθ22)=ρvθ2vmrsin(ϕ)2τbvθsin(α)\begin{gather} \rho v_m \frac{d}{dm}\left(\frac{v_m^2}{2}\right) + v_m\frac{dp}{dm} = \frac{\rho v_\theta^2 v_m}{r} \sin(\phi) - \frac{2 \tau}{b} v_m \cos(\alpha) \\ \rho v_m \frac{d}{dm} \left(\frac{v_\theta^2}{2}\right) = -\frac{\rho v_\theta^2 v_m}{r} \sin(\phi) - \frac{2 \tau}{b} v_\theta \sin(\alpha) \end{gather}

Summing both expressions and recognizing that vm=vcosαv_m = v \cos \alpha and vθ=vsinαv_\theta = v \sin \alpha we find that

vm(dpdm+ρddm(vm22+vθ22))=vm(dpdm+ρvdvdm)=vmdp0dm=(2b)τv v_m \left( \frac{dp}{dm} + \rho \frac{d}{dm}\left(\frac{v_m^2}{2} +\frac{v_\theta^2}{2} \right) \right) = v_m \left( \frac{dp}{dm} + \rho v \frac{dv}{dm} \right) = v_m \frac{dp_0}{dm} = -\left(\frac{2}{b}\right) \tau v

or in other words

dp0dm=2τbcosαdp0ds=2τb \frac{dp_0}{dm} = -\frac{2 \tau}{b \cos \alpha} \to \frac{dp_0}{ds} = -\frac{2 \tau}{b}

That is, loss of stagnation pressure is caused by forces that do not contribute to work in the energy equation. τ\tau in this case. Note that the factor 2/b2/b corresponds to the hydrauling diameter and the 1/cosα1 / \cos \alpha corresponds to the scale factor between the meridional coordinate mm and the length of the streamline ss (that is, the losses are proportional to the length of the streamlines, which makes physical sense)

The loss can be modeled as a summ of several contributions and expressed in terms of the pressure loss coefficient

Y=Δp0p0pY = \frac{\Delta p_0}{p_0-p}

or expressed in differential form

dp0dm=(p0p)dYdm\frac{dp_0}{dm} = (p_0 - p) \frac{dY}{dm}

where Aungier recommends the following split of the loss

dYdm=dYFdm+dYDdm+dYCdm\begin{gather} \frac{dY}{dm} = \frac{dY_{F}}{dm} + \frac{dY_D}{dm} + \frac{dY_C}{dm} \end{gather}

Where

dYFdm=2bCf\frac{dY_{F}}{dm} = \frac{2}{b} C_f

is the loss due to skin friction at the walls

dYDdm=2bD(1E)\frac{dY_{D}}{dm} = \frac{2}{b} D (1-E)

where

D=bAdAdm,Dm=0.4cosαin(binL)0.35D = \frac{b}{A} \frac{dA}{dm}, \qquad D_m = 0.4 \cos \alpha_{in} \left( \frac{b_{in}}{L}\right)^{0.35}
E={1,D010.2(DDm)2,0<D<Dm0.8DmD,DDmE = \begin{cases} 1, & D \le 0 \\ 1 - 0.2 \left( \frac{D}{D_m} \right)^{2}, & 0 < D < D_m \\ 0.8 \sqrt{\frac{D_m}{D}}, & D \ge D_m \end{cases}

And YCY_C the loss due passage curvature, which si given by:

dYFdm=2bbκ26\frac{dY_{F}}{dm} = \frac{2}{b} \frac{b \, \kappa}{26}

where κ\kappa is the local curvature of the streamline

The term II, as suggested by Aungier describes the additional losses due to diffusion (flow deceleration due to area increase) and curvature of the streamwise channel.

Thoughts for the paper

If there is time, we can explain the model for the 90--degree bend including first the flow equations explaining that they where modified for p-h function calls (in agremeent with look-up tables)

Then explain the geometry.

Simply phi=90 and m=r for the case of radial

for the 90-degree bent we have to define the equations of the NURBS

explain the geometric construction of the shape

3-degree bs-pline for the midline defined by parameters

Also a b-spline (linear) for the thickness distribution.

Explain the equations for the construction and arclength/coordinate reparametrization in an appendix?